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german
3 years ago
8

Cuantas moléculas de K² Cr ² 0⁷ hay en 0.5 dm³ de una solución de dicromoto de potasios de 30% en p/p y una densidad relativo de

1.15? (Coronel R., 2018)
Chemistry
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

El número de moléculas de K₂Cr₂O₇ presentes en la solución es de aproximadamente 3,52058347 × 10²³ moléculas

Explanation:

Los parámetros dados de la solución de dicromato de potasio son;

El volumen de la solución, V = 0,5 dm³ = 0,0005 m³

El porcentaje en masa de dicromato de potasio en la solución = 30% p / p

La densidad relativa de la solución, RD = 1,15

Por lo tanto, la masa de dicromato de potasio en la solución, m = 30% de la masa de la solución.

El RD de la solución = (La densidad de la solución) / (La densidad del agua)

La densidad del agua = 997 kg / m³

Por lo tanto, tenemos;

RD = 1,15 = (La densidad de la solución) / (997 kg / m³)

La densidad de la solución, ρ = 1,15 × 997 kg / m³ = 1,146,55 kg / m³

La masa de la solución, m = ρ × V

∴ m = 1.146,55 kg / m³ × 0.0005 m³ = 0,573275 kg

La masa de dicromato de potasio en la solución, m₁ = 0.3 × m

∴ m₁ = 0.3 × 0.573275 kg = 0.1719825 kg = 171.9825 g

La masa de dicromato de potasio en la solución, m₁ = 171,9825 g

La masa molar del dicromato de potasio, K₂Cr₂O₇ = 294,185 g / mol

El número de moles de dicromato de potasio, K₂Cr₂O₇ en la solución, n = (Masa, m) / (Masa molar)

∴ n = 171,9825 g / (294,185 g / mol) = 0,584606625 moles

∴ El número de moléculas de K₂Cr₂O₇ presentes en la solución = n ×

Dónde;

= Número de Avogadro = 6.0221409 × 10²³ mol⁻¹

Por lo tanto, tenemos;

El número de moléculas de K₂Cr₂O₇ presentes en la solución = 0.584606625 moles × 6.0221409 × 10²³ mol⁻¹ = 3.52058347 × 10²³ moléculas

El número de moléculas de K₂Cr₂O₇ presentes en la solución ≈ 3.52058347 × 10²³ moléculas

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